package org.lep.leetcode.symmetrictree;

import java.util.Stack;

/**
 * Source : https://oj.leetcode.com/problems/symmetric-tree/
 *
 * Created by lverpeng on 2017/8/11.
 *
 * Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
 *
 * For example, this binary tree is symmetric:
 *
 *     1
 *    / \
 *   2   2
 *  / \ / \
 * 3  4 4  3
 *
 * But the following is not:
 *
 *     1
 *    / \
 *   2   2
 *    \   \
 *    3    3
 *
 * Note:
 * Bonus points if you could solve it both recursively and iteratively.
 */
public class SymmetricTree {

    /**
     * 判断一棵树是否是镜像对称的
     *
     * 类比判断两棵树是否相同，将一棵树leftNode，rightNode看做两棵树，判断两棵树是否是镜像对称的
     *
     * 根节点都为空，true
     * 只有一个根节点为空，false
     * 两个根节点都不为空且相同，递归判断两个子节点
     *
     * @param left
     * @param right
     * @return
     */
    public boolean isSymmetric (TreeNode left, TreeNode right) {
        if (left == null && right == null) {
            return true;
        }
        if ((left == null && right != null) || (left != null && right == null)) {
            return false;
        }
        if (left.value != right.value) {
            return false;
        }
        return isSymmetric(left.leftChild, right.rightChild) && isSymmetric(left.rightChild, right.leftChild);
    }


    /**
     * 使用递归判断是否是镜像对称的
     *
     * 借助栈实现，
     *
     * @param left
     * @param right
     * @return
     */
    public boolean isSymmetricByIterator (TreeNode left, TreeNode right) {
        Stack<TreeNode> leftStack = new Stack<TreeNode>();
        Stack<TreeNode> rightStack = new Stack<TreeNode>();
        leftStack.push(left);
        rightStack.push(right);
        while (leftStack.size() > 0 && rightStack.size() > 0) {
            TreeNode leftNode = leftStack.pop();
            TreeNode rightNode = rightStack.pop();
            if (leftNode == null && rightNode == null) {
                continue;
            }
            if ((leftNode == null && rightNode != null) || (leftNode != null && rightNode == null)) {
                return false;
            }
            if (leftNode.value != rightNode.value) {
                return false;
            }
            leftStack.push(leftNode.leftChild);
            leftStack.push(leftNode.rightChild);
            rightStack.push(rightNode.rightChild);
            rightStack.push(rightNode.leftChild);
        }
        return true;
    }


    public TreeNode createTree (char[] treeArr) {
        TreeNode[] tree = new TreeNode[treeArr.length];
        for (int i = 0; i < treeArr.length; i++) {
            if (treeArr[i] == '#') {
                tree[i] = null;
                continue;
            }
            tree[i] = new TreeNode(treeArr[i]-'0');
        }
        int pos = 0;
        for (int i = 0; i < treeArr.length && pos < treeArr.length-1; i++) {
            if (tree[i] != null) {
                tree[i].leftChild = tree[++pos];
                if (pos < treeArr.length-1) {
                    tree[i].rightChild = tree[++pos];
                }
            }
        }
        return tree[0];
    }


    private class TreeNode {
        TreeNode leftChild;
        TreeNode rightChild;
        int value;

        public TreeNode(int value) {
            this.value = value;
        }

        public TreeNode() {
        }
    }


    public static void main(String[] args) {
        SymmetricTree symmetricTree = new SymmetricTree();
        char[] treeArr1 = new char[]{'1','2','2','3','4','4','3'};
        char[] treeArr2 = new char[]{'1','2','2','#','4','#','4'};

        TreeNode tree1 = symmetricTree.createTree(treeArr1);
        TreeNode tree2 = symmetricTree.createTree(treeArr2);


        System.out.println(symmetricTree.isSymmetric(tree1.leftChild, tree1.rightChild) + "----true");
        System.out.println(symmetricTree.isSymmetricByIterator(tree1.leftChild, tree1.rightChild) + "----true");


        System.out.println(symmetricTree.isSymmetric(tree2.leftChild, tree2.rightChild) + "----false");
        System.out.println(symmetricTree.isSymmetricByIterator(tree2.leftChild, tree2.rightChild) + "----false");
    }

}
